| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2013 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Sum or combination of independent binomial values |
| Difficulty | Moderate -0.8 This is a straightforward binomial distribution question requiring only standard formula application: E(X) = np, direct use of P(X=k) formula, and recognizing part (iii) as another binomial problem. All parts are routine textbook exercises with no problem-solving insight needed, making it easier than average but not trivial due to the multi-step nature of parts (ii) and (iii). |
| Spec | 5.02c Linear coding: effects on mean and variance5.02d Binomial: mean np and variance np(1-p) |
| Answer | Marks |
|---|---|
| 1.25 oe | B1, [1] |
| Answer | Marks | Guidance |
|---|---|---|
| 0.8965 − 0.6328 = 0.264 (3 sf) | M1, A1, [2] | \(^5C_3(\frac{1}{4})^3(\frac{1}{4})^2\) = \(\frac{135}{512}\) or 0.264 (3 sf) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer which rounds to 0.244 | M1, M1, M1, A1, [4] | \(((\frac{3}{4})^5)^2\) or \((1024)^2\) or \((\frac{243}{1024})^2\) or \((\frac{3}{4})^{10}\) oe (= \(\frac{59049}{1048576}\)); \((\frac{3}{4})^5×5(\frac{1}{4})(\frac{1}{4})\) or \(\frac{243}{1024} × \frac{405}{1024}\) or \(5(\frac{1}{4})^2(\frac{1}{4})\) (= \(\frac{98415}{1048576}\)); \(2x\)(attempt P(1, 0) alone), (NOT \(2×(P(1,0) + P(0,0))\)). If P(sum ≤ 2), all three M−mks are available, but for 3rd M1, must be 2×(P(1,0)+P(2,0)) only |
| Answer | Marks | Guidance |
|---|---|---|
| Use of 0.2637 or 0.264; \(^{10}C_3×(1 −\text{'0.2637'})^7 × \text{'0.2637'}^3\) = 0.258 (3 sf) | M1, M1, A1, [3] | or their (i)(b); ft (i)(b); allow ft their (ii) for this M1 only |
If incorrect $p$ used consistently in all parts of qu 5, no mks in 5(i)(a) & (b) but can score M−marks in (ii) and (iii).
## (i)(a)
1.25 oe | B1, [1]
## (i)(b)
0.8965 − 0.6328 = 0.264 (3 sf) | M1, A1, [2] | $^5C_3(\frac{1}{4})^3(\frac{1}{4})^2$ = $\frac{135}{512}$ or 0.264 (3 sf) | Answer which rounds to 0.264
## (ii)
Answer which rounds to 0.244 | M1, M1, M1, A1, [4] | $((\frac{3}{4})^5)^2$ or $(1024)^2$ or $(\frac{243}{1024})^2$ or $(\frac{3}{4})^{10}$ oe (= $\frac{59049}{1048576}$); $(\frac{3}{4})^5×5(\frac{1}{4})(\frac{1}{4})$ or $\frac{243}{1024} × \frac{405}{1024}$ or $5(\frac{1}{4})^2(\frac{1}{4})$ (= $\frac{98415}{1048576}$); $2x$(attempt P(1, 0) alone), (NOT $2×(P(1,0) + P(0,0))$). If P(sum ≤ 2), all three M−mks are available, but for 3rd M1, must be 2×(P(1,0)+P(2,0)) only | B(10, 0.25) seen or implied; M1; Table or formula with $n = 10$ used; M1; P(X ≤ 1) from table or $(\frac{3}{4})^{10} + 10(\frac{3}{4})^9×(\frac{1}{4})$; M1; 0.244 (3 sf); A1; P(X < 2) = 0.526 from table $n = 10$; M1M1M1A0; SC P(X = 2) answer 0.282: B1
## (iii)
Use of 0.2637 or 0.264; $^{10}C_3×(1 −\text{'0.2637'})^7 × \text{'0.2637'}^3$ = 0.258 (3 sf) | M1, M1, A1, [3] | or their (i)(b); ft (i)(b); allow ft their (ii) for this M1 only | SC allow $^{10}C_3×(1 −\text{'0.282'})^7×\text{'0.282'}^3$ (0.282 comes from P(3 totals = 2)); M0M1; Correct ans, no working: M1M1A1
---A random variable $X$ has the distribution B$(5, \frac{1}{4})$.
\begin{enumerate}[label=(\roman*)]
\item Find
\begin{enumerate}[label=(\alph*)]
\item E($X$), [1]
\item P($X = 2$). [2]
\end{enumerate}
\item Two values of $X$ are chosen at random. Find the probability that their sum is less than 2. [4]
\item 10 values of $X$ are chosen at random. Use an appropriate formula to find the probability that exactly 3 of these values are 2s. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR S1 2013 Q5 [10]}}